\(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) [225]

Optimal result

Integrand size = 23, antiderivative size = 107 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {1}{2} \left (2 a^2 A+A b^2+2 a b B\right ) x+\frac {2 \left (3 a A b+a^2 B+b^2 B\right ) \sin (c+d x)}{3 d}+\frac {b (3 A b+2 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \]

[Out]

1/2*(2*A*a^2+A*b^2+2*B*a*b)*x+2/3*(3*A*a*b+B*a^2+B*b^2)*sin(d*x+c)/d+1/6*b*(3*A*b+2*B*a)*cos(d*x+c)*sin(d*x+c)
/d+1/3*B*(a+b*cos(d*x+c))^2*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2832, 2813} \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {2 \left (a^2 B+3 a A b+b^2 B\right ) \sin (c+d x)}{3 d}+\frac {1}{2} x \left (2 a^2 A+2 a b B+A b^2\right )+\frac {b (2 a B+3 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {B \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

[In]

Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x]),x]

[Out]

((2*a^2*A + A*b^2 + 2*a*b*B)*x)/2 + (2*(3*a*A*b + a^2*B + b^2*B)*Sin[c + d*x])/(3*d) + (b*(3*A*b + 2*a*B)*Cos[
c + d*x]*Sin[c + d*x])/(6*d) + (B*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d)

Rule 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2832

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Sim
p[b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rubi steps \begin{align*} \text {integral}& = \frac {B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} \int (a+b \cos (c+d x)) (3 a A+2 b B+(3 A b+2 a B) \cos (c+d x)) \, dx \\ & = \frac {1}{2} \left (2 a^2 A+A b^2+2 a b B\right ) x+\frac {2 \left (3 a A b+a^2 B+b^2 B\right ) \sin (c+d x)}{3 d}+\frac {b (3 A b+2 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.84 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {6 \left (2 a^2 A+A b^2+2 a b B\right ) (c+d x)+3 \left (8 a A b+4 a^2 B+3 b^2 B\right ) \sin (c+d x)+3 b (A b+2 a B) \sin (2 (c+d x))+b^2 B \sin (3 (c+d x))}{12 d} \]

[In]

Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x]),x]

[Out]

(6*(2*a^2*A + A*b^2 + 2*a*b*B)*(c + d*x) + 3*(8*a*A*b + 4*a^2*B + 3*b^2*B)*Sin[c + d*x] + 3*b*(A*b + 2*a*B)*Si
n[2*(c + d*x)] + b^2*B*Sin[3*(c + d*x)])/(12*d)

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(\frac {\left (3 A \,b^{2}+6 B a b \right ) \sin \left (2 d x +2 c \right )+B \sin \left (3 d x +3 c \right ) b^{2}+\left (24 A a b +12 B \,a^{2}+9 B \,b^{2}\right ) \sin \left (d x +c \right )+12 x \left (A \,a^{2}+\frac {1}{2} A \,b^{2}+B a b \right ) d}{12 d}\)	\(88\)
parts	\(a^{2} x A +\frac {\left (A \,b^{2}+2 B a b \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (2 A a b +B \,a^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {B \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(91\)
derivativedivides	\(\frac {\frac {B \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \sin \left (d x +c \right ) a b +B \sin \left (d x +c \right ) a^{2}+A \,a^{2} \left (d x +c \right )}{d}\)	\(114\)
default	\(\frac {\frac {B \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \sin \left (d x +c \right ) a b +B \sin \left (d x +c \right ) a^{2}+A \,a^{2} \left (d x +c \right )}{d}\)	\(114\)
risch	\(a^{2} x A +\frac {x A \,b^{2}}{2}+x B a b +\frac {2 \sin \left (d x +c \right ) A a b}{d}+\frac {\sin \left (d x +c \right ) B \,a^{2}}{d}+\frac {3 b^{2} B \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (3 d x +3 c \right ) B \,b^{2}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a b}{2 d}\)	\(116\)
norman	\(\frac {\left (A \,a^{2}+\frac {1}{2} A \,b^{2}+B a b \right ) x +\left (A \,a^{2}+\frac {1}{2} A \,b^{2}+B a b \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,a^{2}+\frac {3}{2} A \,b^{2}+3 B a b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,a^{2}+\frac {3}{2} A \,b^{2}+3 B a b \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (4 A a b -A \,b^{2}+2 B \,a^{2}-2 B a b +2 B \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (4 A a b +A \,b^{2}+2 B \,a^{2}+2 B a b +2 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 \left (6 A a b +3 B \,a^{2}+B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(245\)

[In]

int((a+cos(d*x+c)*b)^2*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/12*((3*A*b^2+6*B*a*b)*sin(2*d*x+2*c)+B*sin(3*d*x+3*c)*b^2+(24*A*a*b+12*B*a^2+9*B*b^2)*sin(d*x+c)+12*x*(A*a^2
+1/2*A*b^2+B*a*b)*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.79 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {3 \, {\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} d x + {\left (2 \, B b^{2} \cos \left (d x + c\right )^{2} + 6 \, B a^{2} + 12 \, A a b + 4 \, B b^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*(2*A*a^2 + 2*B*a*b + A*b^2)*d*x + (2*B*b^2*cos(d*x + c)^2 + 6*B*a^2 + 12*A*a*b + 4*B*b^2 + 3*(2*B*a*b +
A*b^2)*cos(d*x + c))*sin(d*x + c))/d

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.86 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\begin {cases} A a^{2} x + \frac {2 A a b \sin {\left (c + d x \right )}}{d} + \frac {A b^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A b^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {A b^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {B a^{2} \sin {\left (c + d x \right )}}{d} + B a b x \sin ^{2}{\left (c + d x \right )} + B a b x \cos ^{2}{\left (c + d x \right )} + \frac {B a b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {2 B b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right )^{2} & \text {otherwise} \end {cases} \]

[In]

integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((A*a**2*x + 2*A*a*b*sin(c + d*x)/d + A*b**2*x*sin(c + d*x)**2/2 + A*b**2*x*cos(c + d*x)**2/2 + A*b**
2*sin(c + d*x)*cos(c + d*x)/(2*d) + B*a**2*sin(c + d*x)/d + B*a*b*x*sin(c + d*x)**2 + B*a*b*x*cos(c + d*x)**2
+ B*a*b*sin(c + d*x)*cos(c + d*x)/d + 2*B*b**2*sin(c + d*x)**3/(3*d) + B*b**2*sin(c + d*x)*cos(c + d*x)**2/d,
Ne(d, 0)), (x*(A + B*cos(c))*(a + b*cos(c))**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.01 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {12 \, {\left (d x + c\right )} A a^{2} + 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{2} + 12 \, B a^{2} \sin \left (d x + c\right ) + 24 \, A a b \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*A*a^2 + 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a*b + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^2
- 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b^2 + 12*B*a^2*sin(d*x + c) + 24*A*a*b*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.87 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=\frac {B b^{2} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {1}{2} \, {\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} x + \frac {{\left (2 \, B a b + A b^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \]

[In]

integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/12*B*b^2*sin(3*d*x + 3*c)/d + 1/2*(2*A*a^2 + 2*B*a*b + A*b^2)*x + 1/4*(2*B*a*b + A*b^2)*sin(2*d*x + 2*c)/d +
1/4*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*sin(d*x + c)/d

Mupad [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.07 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx=A\,a^2\,x+\frac {A\,b^2\,x}{2}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,B\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+B\,a\,b\,x+\frac {A\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {2\,A\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \]

[In]

int((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^2,x)

[Out]

A*a^2*x + (A*b^2*x)/2 + (B*a^2*sin(c + d*x))/d + (3*B*b^2*sin(c + d*x))/(4*d) + B*a*b*x + (A*b^2*sin(2*c + 2*d
*x))/(4*d) + (B*b^2*sin(3*c + 3*d*x))/(12*d) + (2*A*a*b*sin(c + d*x))/d + (B*a*b*sin(2*c + 2*d*x))/(2*d)